Results 1 to 3 of 3

Thread: evaluating solutions to trig equation??

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    175

    evaluating solutions to trig equation??

    $\displaystyle
    \begin{array}{l}
    {\rm Trying to find all solutions to interval inclusive of - pi to + pi for } \\
    2 + \cos (2x) = 3\cos (x) \\
    {\rm What I have done is this:} \\
    2 + [\cos ^2 (x) - \sin ^2 (x)] = 3\cos (x) \\
    2 + \cos ^2 (x) - \sin ^2 (x) = 3\cos (x) \\
    2 + \cos ^2 (x) - 3\cos (x) = \sin ^2 (x) \\
    \end{array}$

    $\displaystyle
    \begin{array}{l}
    2 + \cos ^2 (x) - 3\cos (x) = 1 - \cos ^2 (x) \\
    2 + 2\cos ^2 (x) - 3\cos (x) = 1 \\
    2\cos ^2 (x) - 3\cos (x) + 1 = 0 \\
    (2\cos (x) - 2)(\cos (x) - 1) = 0 \\
    {\rm So then }2\cos (x) = 2{\rm and }\cos (x) = 1 \\
    {\rm Which gives one root of cos(x) = 1} \\
    \end{array}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by Craka View Post
    $\displaystyle
    \begin{array}{l}
    {\rm Trying to find all solutions to interval inclusive of - pi to + pi for } \\
    2 + \cos (2x) = 3\cos (x) \\
    {\rm What I have done is this:} \\
    2 + [\cos ^2 (x) - \sin ^2 (x)] = 3\cos (x) \\
    2 + \cos ^2 (x) - \sin ^2 (x) = 3\cos (x) \\
    2 + \cos ^2 (x) - 3\cos (x) = \sin ^2 (x) \\
    \end{array}$

    $\displaystyle
    \begin{array}{l}
    2 + \cos ^2 (x) - 3\cos (x) = 1 - \cos ^2 (x) \\
    2 + 2\cos ^2 (x) - 3\cos (x) = 1 \\
    2\cos ^2 (x) - 3\cos (x) + 1 = 0 \\
    (2\cos (x) - 2)(\cos (x) - 1) = 0 \\
    {\rm So then }2\cos (x) = 2{\rm and }\cos (x) = 1 \\
    {\rm Which gives one root of cos(x) = 1} \\
    \end{array}$
    Something went wrong in your factorisation

    $\displaystyle (2 \cos(x)-2)(\cos(x)-1)=2(\cos(x)-1)^2=2 \cos^2(x)-{\color{red}4}\cos(x)+{\color{red}2}$

    $\displaystyle 2 \cos^2(x)-3\cos(x)+1=(2 \cos(x)-{\color{magenta}1})(\cos(x)-1)=0$

    This gives $\displaystyle \cos(x)=\frac 12$ or $\displaystyle \cos(x)=1$.
    Last edited by Moo; Aug 22nd 2008 at 07:45 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, Craka!

    There are three forms for $\displaystyle \cos2\theta$

    . . $\displaystyle \cos2\theta \;=\;\begin{Bmatrix}\cos^2\!\theta - \sin^2\!\theta \\ 2\cos^2\!\theta - 1 \\ 1 - 2\sin^2\!\theta \end{Bmatrix} $


    The last two can be derived from the first,
    but I've found it expedient to memorize all three.


    Find all solutions on $\displaystyle [-\pi,\:+\pi]\!:\;\;2 + \cos (2x) \:=\:3\cos (x)$

    We have: . $\displaystyle 2 + \cos2x \:=\:3\cos x$

    . . . . . $\displaystyle 2 + \overbrace{2\cos^2\!x - 1} \:=\:3\cos x$

    . . $\displaystyle 2\cos^2\!x - 3\cos x + 1 \:=\:0$

    .$\displaystyle (\cos x - 1)(2\cos x - 1) \:=\:0$


    Therefore: . $\displaystyle \begin{array}{ccccc}\cos x - 1 \:=\:0 & \Rightarrow & \cos x \:=\:1 & \Rightarrow &\boxed{x \:=\:0} \\ \\[-3mm]
    2\cos x - 1 \:=\:0 & \Rightarrow & \cos x \:=\:\frac{1}{2} & \Rightarrow & \boxed{x \:=\:\pm\frac{\pi}{3}} \end{array}$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. All solutions for trig equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Sep 13th 2011, 05:52 AM
  2. Finding all solutions to the trig equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Dec 11th 2010, 10:00 PM
  3. trig equation - all solutions
    Posted in the Trigonometry Forum
    Replies: 11
    Last Post: Jan 21st 2010, 01:23 PM
  4. Trig equation (and its solutions)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Sep 25th 2008, 08:54 AM
  5. Trig equation - graphing to show solutions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 1st 2008, 08:31 AM

Search Tags


/mathhelpforum @mathhelpforum